What is the general solution for arctan(1/(x-1))+arctan(2/(x+1))= pi/4 ?

The general solution for arctan(1/(x-1))+arctan(2/(x+1))= pi/4 is x=(3+sqrt(17))/2 ,x=(3-sqrt(17))/2

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Popular Trigonometry >

arctan(1/(x-1))+arctan(2/(x+1))= pi/4

Solution

arctan (\frac{1}{x - 1}) + arctan (\frac{2}{x + 1}) = \frac{π}{4}

Solution

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

Solution steps

arctan (\frac{1}{x - 1}) + arctan (\frac{2}{x + 1}) = \frac{π}{4}

Rewrite using trig identities

arctan (\frac{1}{x - 1}) + arctan (\frac{2}{x + 1})

Use the Sum to Product identity:

arctan (s) + arctan (t) = arctan (\frac{s + t}{1 - s t})

= arctan (\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}})

arctan (\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}}) = \frac{π}{4}

Apply trig inverse properties

arctan (\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}}) = \frac{π}{4}

arctan (x) = a \Rightarrow x = tan (a)

\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}} = tan (\frac{π}{4})

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

Use the following trivial identity:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

= 1

\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}} = 1

\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}} = 1

Solve

\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}} = 1 : x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}} = 1

Simplify

\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}} : \frac{3 x - 1}{x ^{2} - 3}

\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}}

\frac{1}{x - 1} \cdot \frac{2}{x + 1} = \frac{2}{( x - 1 ) ( x + 1 )}

\frac{1}{x - 1} \cdot \frac{2}{x + 1}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot 2}{( x - 1 ) ( x + 1 )}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{( x - 1 ) ( x + 1 )}

= \frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{2}{( x - 1 ) ( x + 1 )}}

Join

\frac{1}{x - 1} + \frac{2}{x + 1} : \frac{3 x - 1}{( x - 1 ) ( x + 1 )}

\frac{1}{x - 1} + \frac{2}{x + 1}

Least Common Multiplier of

x - 1, x + 1 : (x - 1) (x + 1)

x - 1, x + 1

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

x - 1

x + 1

= (x - 1) (x + 1)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

(x - 1) (x + 1)

For

\frac{1}{x - 1} :

multiply the denominator and numerator by

x + 1

\frac{1}{x - 1} = \frac{1 \cdot ( x + 1 )}{( x - 1 ) ( x + 1 )} = \frac{x + 1}{( x - 1 ) ( x + 1 )}

For

\frac{2}{x + 1} :

multiply the denominator and numerator by

x - 1

\frac{2}{x + 1} = \frac{2 ( x - 1 )}{( x + 1 ) ( x - 1 )}

= \frac{x + 1}{( x - 1 ) ( x + 1 )} + \frac{2 ( x - 1 )}{( x + 1 ) ( x - 1 )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{x + 1 + 2 ( x - 1 )}{( x - 1 ) ( x + 1 )}

Expand

x + 1 + 2 (x - 1) : 3 x - 1

x + 1 + 2 (x - 1)

Expand

2 (x - 1) : 2 x - 2

2 (x - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = x, c = 1

= 2 x - 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2 x - 2

= x + 1 + 2 x - 2

Simplify

x + 1 + 2 x - 2 : 3 x - 1

x + 1 + 2 x - 2

Group like terms

= x + 2 x + 1 - 2

Add similar elements:

x + 2 x = 3 x

= 3 x + 1 - 2

Add/Subtract the numbers:

1 - 2 = - 1

= 3 x - 1

= 3 x - 1

= \frac{3 x - 1}{( x - 1 ) ( x + 1 )}

= \frac{\frac{3 x - 1}{( x - 1 ) ( x + 1 )}}{1 - \frac{2}{( x - 1 ) ( x + 1 )}}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{3 x - 1}{( x - 1 ) ( x + 1 ) ( 1 - \frac{2}{( x - 1 ) ( x + 1 )} )}

Join

1 - \frac{2}{( x - 1 ) ( x + 1 )} : \frac{x ^{2} - 3}{( x - 1 ) ( x + 1 )}

1 - \frac{2}{( x - 1 ) ( x + 1 )}

Convert element to fraction:

1 = \frac{1 ( x - 1 ) ( x + 1 )}{( x - 1 ) ( x + 1 )}

= \frac{1 \cdot ( x - 1 ) ( x + 1 )}{( x - 1 ) ( x + 1 )} - \frac{2}{( x - 1 ) ( x + 1 )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot ( x - 1 ) ( x + 1 ) - 2}{( x - 1 ) ( x + 1 )}

Multiply:

1 \cdot (x - 1) = (x - 1)

= \frac{( x - 1 ) ( x + 1 ) - 2}{( x - 1 ) ( x + 1 )}

Expand

(x - 1) (x + 1) - 2 : x^{2} - 3

(x - 1) (x + 1) - 2

Expand

(x - 1) (x + 1) : x^{2} - 1

(x - 1) (x + 1)

Apply Difference of Two Squares Formula:

(a - b) (a + b) = a^{2} - b^{2}

a = x, b = 1

= x^{2} - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= x^{2} - 1

= x^{2} - 1 - 2

Subtract the numbers:

- 1 - 2 = - 3

= x^{2} - 3

= \frac{x ^{2} - 3}{( x - 1 ) ( x + 1 )}

= \frac{3 x - 1}{\frac{x ^{2} - 3}{( x - 1 ) ( x + 1 )} ( x - 1 ) ( x + 1 )}

Multiply

(x - 1) (x + 1) \frac{x ^{2} - 3}{( x - 1 ) ( x + 1 )} : x^{2} - 3

(x - 1) (x + 1) \frac{x ^{2} - 3}{( x - 1 ) ( x + 1 )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( x ^{2} - 3 ) ( x - 1 ) ( x + 1 )}{( x - 1 ) ( x + 1 )}

Cancel the common factor:

x - 1

= \frac{( x ^{2} - 3 ) ( x + 1 )}{x + 1}

Cancel the common factor:

x + 1

= x^{2} - 3

= \frac{3 x - 1}{x ^{2} - 3}

\frac{3 x - 1}{x ^{2} - 3} = 1

Multiply both sides by

x^{2} - 3

\frac{3 x - 1}{x ^{2} - 3} = 1

Multiply both sides by

x^{2} - 3

\frac{3 x - 1}{x ^{2} - 3} (x^{2} - 3) = 1 \cdot (x^{2} - 3)

Simplify

\frac{3 x - 1}{x ^{2} - 3} (x^{2} - 3) = 1 \cdot (x^{2} - 3)

Simplify

\frac{3 x - 1}{x ^{2} - 3} (x^{2} - 3) : 3 x - 1

\frac{3 x - 1}{x ^{2} - 3} (x^{2} - 3)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 3 x - 1 ) ( x ^{2} - 3 )}{x ^{2} - 3}

Cancel the common factor:

x^{2} - 3

= 3 x - 1

Simplify

1 \cdot (x^{2} - 3) : x^{2} - 3

1 \cdot (x^{2} - 3)

Multiply:

1 \cdot (x^{2} - 3) = (x^{2} - 3)

= (x^{2} - 3)

Remove parentheses:

(a) = a

= x^{2} - 3

3 x - 1 = x^{2} - 3

3 x - 1 = x^{2} - 3

3 x - 1 = x^{2} - 3

Solve

3 x - 1 = x^{2} - 3 : x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

3 x - 1 = x^{2} - 3

Switch sides

x^{2} - 3 = 3 x - 1

Move

1

to the left side

x^{2} - 3 = 3 x - 1

Add

1

to both sides

x^{2} - 3 + 1 = 3 x - 1 + 1

Simplify

x^{2} - 2 = 3 x

x^{2} - 2 = 3 x

Move

3 x

to the left side

x^{2} - 2 = 3 x

Subtract

3 x

from both sides

x^{2} - 2 - 3 x = 3 x - 3 x

Simplify

x^{2} - 2 - 3 x = 0

x^{2} - 2 - 3 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 3 x - 2 = 0

Solve with the quadratic formula

x^{2} - 3 x - 2 = 0

Quadratic Equation Formula:

For

a = 1, b = - 3, c = - 2

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 2) = 17

(- 3)^{2} - 4 \cdot 1 \cdot (- 2)

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 1 \cdot 2

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 3^{2} + 8

3^{2} = 9

= 9 + 8

Add the numbers:

9 + 8 = 17

= 17

x_{1, 2} = \frac{- ( - 3 ) \pm 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 17}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 17}{2 \cdot 1}

x = \frac{- ( - 3 ) + 17}{2 \cdot 1} : \frac{3 + 17}{2}

\frac{- ( - 3 ) + 17}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 + 17}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{3 + 17}{2}

x = \frac{- ( - 3 ) - 17}{2 \cdot 1} : \frac{3 - 17}{2}

\frac{- ( - 3 ) - 17}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 - 17}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{3 - 17}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

Verify Solutions

Find undefined (singularity) points:

x = 3, x = - 3, x = 1, x = - 1

Take the denominator(s) of

\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}}

and compare to zero

Solve

1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1} = 0 : x = 3, x = - 3

1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1} = 0

Simplify

- \frac{1}{x - 1} \cdot \frac{2}{x + 1} : - \frac{2}{( x - 1 ) ( x + 1 )}

- \frac{1}{x - 1} \cdot \frac{2}{x + 1}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= - \frac{1 \cdot 2}{( x - 1 ) ( x + 1 )}

Multiply the numbers:

1 \cdot 2 = 2

= - \frac{2}{( x - 1 ) ( x + 1 )}

1 - \frac{2}{( x - 1 ) ( x + 1 )} = 0

Multiply both sides by

(x - 1) (x + 1)

1 - \frac{2}{( x - 1 ) ( x + 1 )} = 0

Multiply both sides by

(x - 1) (x + 1)

1 \cdot (x - 1) (x + 1) - \frac{2}{( x - 1 ) ( x + 1 )} (x - 1) (x + 1) = 0 \cdot (x - 1) (x + 1)

Simplify

1 \cdot (x - 1) (x + 1) - \frac{2}{( x - 1 ) ( x + 1 )} (x - 1) (x + 1) = 0 \cdot (x - 1) (x + 1)

Simplify

1 \cdot (x - 1) (x + 1) : (x - 1) (x + 1)

1 \cdot (x - 1) (x + 1)

Multiply:

1 \cdot (x - 1) = (x - 1)

= (x - 1) (x + 1)

Simplify

- \frac{2}{( x - 1 ) ( x + 1 )} (x - 1) (x + 1) : - 2

- \frac{2}{( x - 1 ) ( x + 1 )} (x - 1) (x + 1)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{2 ( x - 1 ) ( x + 1 )}{( x - 1 ) ( x + 1 )}

Cancel the common factor:

x - 1

= - \frac{2 ( x + 1 )}{x + 1}

Cancel the common factor:

x + 1

= - 2

Simplify

0 \cdot (x - 1) (x + 1) : 0

0 \cdot (x - 1) (x + 1)

Apply rule

0 \cdot a = 0

= 0

(x - 1) (x + 1) - 2 = 0

(x - 1) (x + 1) - 2 = 0

(x - 1) (x + 1) - 2 = 0

Solve

(x - 1) (x + 1) - 2 = 0 : x = 3, x = - 3

(x - 1) (x + 1) - 2 = 0

Expand

(x - 1) (x + 1) - 2 : x^{2} - 3

(x - 1) (x + 1) - 2

Expand

(x - 1) (x + 1) : x^{2} - 1

(x - 1) (x + 1)

Apply Difference of Two Squares Formula:

(a - b) (a + b) = a^{2} - b^{2}

a = x, b = 1

= x^{2} - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= x^{2} - 1

= x^{2} - 1 - 2

Subtract the numbers:

- 1 - 2 = - 3

= x^{2} - 3

x^{2} - 3 = 0

Solve with the quadratic formula

x^{2} - 3 = 0

Quadratic Equation Formula:

For

a = 1, b = 0, c = - 3

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

0^{2} - 4 \cdot 1 \cdot (- 3) = 23

0^{2} - 4 \cdot 1 \cdot (- 3)

Apply rule

0^{a} = 0

0^{2} = 0

= 0 - 4 \cdot 1 \cdot (- 3)

Apply rule

- (- a) = a

= 0 + 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 0 + 12

Add the numbers:

0 + 12 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

x_{1, 2} = \frac{- 0 \pm 2 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 0 + 2 3}{2 \cdot 1}, x_{2} = \frac{- 0 - 2 3}{2 \cdot 1}

x = \frac{- 0 + 2 3}{2 \cdot 1} : 3

\frac{- 0 + 2 3}{2 \cdot 1}

- 0 + 23 = 23

= \frac{2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 3}{2}

Divide the numbers:

\frac{2}{2} = 1

= 3

x = \frac{- 0 - 2 3}{2 \cdot 1} : - 3

\frac{- 0 - 2 3}{2 \cdot 1}

- 0 - 23 = - 23

= \frac{- 2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2 3}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 3}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 3

The solutions to the quadratic equation are:

x = 3, x = - 3

x = 3, x = - 3

Verify Solutions

Find undefined (singularity) points:

x = 1, x = - 1

Take the denominator(s) of

1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}

and compare to zero

Solve

x - 1 = 0 : x = 1

x - 1 = 0

Move

1

to the right side

x - 1 = 0

Add

1

to both sides

x - 1 + 1 = 0 + 1

Simplify

x = 1

x = 1

Solve

x + 1 = 0 : x = - 1

x + 1 = 0

Move

1

to the right side

x + 1 = 0

Subtract

1

from both sides

x + 1 - 1 = 0 - 1

Simplify

x = - 1

x = - 1

The following points are undefined

x = 1, x = - 1

Combine undefined points with solutions:

x = 3, x = - 3

Solve

x - 1 = 0 : x = 1

x - 1 = 0

Move

1

to the right side

x - 1 = 0

Add

1

to both sides

x - 1 + 1 = 0 + 1

Simplify

x = 1

x = 1

Solve

x + 1 = 0 : x = - 1

x + 1 = 0

Move

1

to the right side

x + 1 = 0

Subtract

1

from both sides

x + 1 - 1 = 0 - 1

Simplify

x = - 1

x = - 1

The following points are undefined

x = 3, x = - 3, x = 1, x = - 1

Combine undefined points with solutions:

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

arctan (\frac{1}{x - 1}) + arctan (\frac{2}{x + 1}) = \frac{π}{4}

Remove the ones that don't agree with the equation.

Check the solution

\frac{3 + 17}{2} :

True

\frac{3 + 17}{2}

Plug in

n = 1

\frac{3 + 17}{2}

For

arctan (\frac{1}{x - 1}) + arctan (\frac{2}{x + 1}) = \frac{π}{4}

plug in

x = \frac{3 + 17}{2}

arctan (\frac{1}{\frac{3 + 17}{2} - 1}) + arctan (\frac{2}{\frac{3 + 17}{2} + 1}) = \frac{π}{4}

Refine

0.78539 \dots = 0.78539 \dots

\Rightarrow True

Check the solution

\frac{3 - 17}{2} :

True

\frac{3 - 17}{2}

Plug in

n = 1

\frac{3 - 17}{2}

For

arctan (\frac{1}{x - 1}) + arctan (\frac{2}{x + 1}) = \frac{π}{4}

plug in

x = \frac{3 - 17}{2}

arctan (\frac{1}{\frac{3 - 17}{2} - 1}) + arctan (\frac{2}{\frac{3 - 17}{2} + 1}) = \frac{π}{4}

Refine

0.78539 \dots = 0.78539 \dots

\Rightarrow True

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for arctan(1/(x-1))+arctan(2/(x+1))= pi/4 ?
The general solution for arctan(1/(x-1))+arctan(2/(x+1))= pi/4 is x=(3+sqrt(17))/2 ,x=(3-sqrt(17))/2